MEM 255 Introduction to Control Systems: Analyzing Dynamic Response

Transcription

1 MEM 55 Itroductio to Cotrol Systems: Aalyzig Dyamic Respose Harry G. Kwaty Departmet of Mechaical Egieerig & Mechaics Drexel Uiversity

2 Outlie Time domai ad frequecy domai A secod order system Via partial fractio expasio Usig Matlab Time respose parameters Frequecy respose & Bode plots Example: Suspesio System

3 Time Domai vs Frequecy Domai Aalysis state iput u( t ) xt output y( t) System Time domai: take u t to be a 'simple' fuctio usually a impulse or a step. Characterize the output respose, respose graph. () y t, i terms of the time u() t ωt () = ( ω + φ) ( ω) φ( ω) Frequecy respose: Take to be the siusoidsi. Characterize the output respose, y t Bsi t, i terms of B,.

4 d Order System-Factorig the Deomiator Gs () = Gs = = as + as+ a 1 s + bs 1 + b special case s ω + ρs+ ω ω ( s + ωρ± ω 1+ ρ)

5 Step Respose of d Order System Y = + ρ + ω + ρ + ω ( s) = U( s) c c c = ω ω 1 s s s s s ( + ωρ + ω 1+ ρ)( + ωρ ω 1+ ρ) s s s 1 = + + = ( s+ ωρ + ω 1+ ρ) ( s+ ωρ ω 1+ ρ) s ( s + ωρ + ω 1+ ρ) ω lim s ωρ ω 1+ ρ( s + ωρ + ω 1+ ρ)( s+ ωρ ω 1+ ρ) = lim s s c sω + ρs + ω ω 1 1 =, c = 1+ ρ ρ+ 1+ ρ 1+ ρ ρ 1+ ρ 1 = 1 s c c 3 s

6 Step Respose of d Order System y() t = L Y() s c1l = ( s + ωρ 1 ) + ω + ρ () cl + cl 3 s + ωρ ω 1 ρ s + () tω ρ+ 1+ ρ tω ρ 1+ ρ cu t e c u t e u t = s s s For ρ < 1, the roots are complex tω ρ ρ yt () = 1 e cos tω 1 ρ + si tω 1 ρ 1 ρ ()

7 Step Respose Usig MATLAB >>[Y1,T1]=step(1/(s^+.5*s+1),15); >>[Y,T]=step(1/(s^+.5*s+1),15); >>[Y3,T3]=step(1/(s^+.77*s+1),15); >>[Y4,T4]=step(1/(s^+1.*s+1),15); >>[Y5,T5]=step(1/(s^+1.5*s+1),15); >>plot(t1,y1,'r',t,y,'g',t3,y3,'b',t4,y4,'c',t5,y5,'m')

8 Traditioal Parameters, 1 rise time, T r, usually defied as the time to get from 1% to 9% of its ultimate (i.e., fial) value. settlig time, T s, the time at which the trajectory first eters a ε-tolerace of its ultimate value ad remais there (ε is ofte take as % of the ultimate value). peak time, T p, the time at which the trajectory attais its peak value. peak overshoot, OS, the peak or supreme value of the trajectory ordiarily expressed as a percetage of the ultimate value of the trajectory. A overshoot of more tha 3% is ofte cosidered udesirable. A system without overshoot is overdamped ad may be too slow (as measured by rise time ad settlig time).

9 Traditioal Parameters, y p 1.4 Im 1. α degree of stability, decay rate 1/α 1 ε.8.6 ideal regio for closed loop poles θ dampig ratio θ = si 1 Re ρ t T r T p T s Time respose parameters Ideal pole locatios

10 Frequecy Respose, 1 = cosωt y ( t) u t System Trick: replace cos ωt by e jωt recall e = cosωt+ jsiωt () ω () () () superpositio if cos ωt y t, si t y t the jωt 1 jωt e y1 t + jy t

11 Frequecy Respose, Suppose, 1 G( jω ) ( ω ) ( ω) s 1 s c = L = = + jωt 1 Y s G s e K d s s j ω d s s j s j c1 = lim ( s jω) K = K = G( jω) s jω d s s jω d j ( s) Y() s = + y t = ytras t + yss t d( s) s jω Suppose G s trasiet respose y tras () t s = K d s steady state respose ω () () () () () = ( ω ) j t with time, so y t y t G j e ω ss

12 Frequecy Respose, 3 ( ω) For each specified ω, G is a complex umber, write i polar form: ( ω) jφω jωt () = ρ( ω) = ρ( ω) cos ( ω) ρ( ω) G j G ω = G jω e jφ ω ρ ω e write as y t e e e ss ( ωt+ φ( ω) ) ( t ) j si ( t ) ( t ) ( t ) cos si cosωt ρ ω ω + φ ω siωt ρ ω ω + φ ω G j = ρ ω ω + φ ω + ρ ω ω + φ ω jφω = e is called the frequecy trasfer fuctio j

13 Accelerometer Gss=ss(A,B,C,); G=tf(Gss); bode(g) Bode Diagram Magitude (db) Phase (deg) Frequecy (rad/sec)

14 Accelerometer, Output Itegrator Compesated G1=G*(1/s); bode(g1) Bode Diagram -5 Magitude (db) Phase (deg) Frequecy (rad/sec)

15 Auto Suspesio: quarter car model k k 1 m m 1 y( t) x x 1 mx = c x x + k x x k x y t mx = k x x c x x x 1 1 = v 1 1 ( + ) k k k c c k v = x + x v + v + y t m1 m1 m1 m1 m1 x = v k k c c v = x x + v v 1 1 m m m m ()

16 Suspesio g = 3.; m = 1/g; m1 = 1/g; k = 1; k1 = 1*k; c=*.77*sqrt(k/m) A=[ 1 ;(-k1-k)/m1,- c/m1,k/m1,c/m1;,,,1;k/m,c/m, -k/m,-c/m]; B=[;k1/m1;;]; % payload positio C=[ 1 ]; Gss=ss(A,B,C,) bode(gss,{.1,1}) zpk(gss) % payload velocity pause C=[ 1]; Gss=ss(A,B,C,) bode(gss,{.1,1}) zpk(gss) % payload acceleratio pause C=[k/m c/m -k/m -c/m]; Gss=ss(A,B,C,) bode(gss,{.1,1}) zpk(gss)

17 Suspesio: displacemet respose 5 Bode Diagram Magitude (db) Phase (deg) Frequecy (rad/sec)

18 Suspesio: velocity respose 5 Bode Diagram Magitude (db) Phase (deg) Frequecy (rad/sec)

19 Suspesio: acceleratio respose Bode Diagram Magitude (db) Phase (deg) Frequecy (rad/sec)

20 Suspesio: Effect of Dampig Ratio (velocity) 5 Bode Diagram Magitude (db) Phase (deg) Frequecy (rad/sec)

21 Suspesio: road disturbace x

22 Road Disturbace Calculatios := H8ê 1L vtd,t,sdd := EvaluateA ^H ktsle k= := s,.1, 1D

23 Summary Time Respose Respose to uit step iput Time respose parameters, % overshoot, settlig time Pole locatios Frequecy Respose Respose to siusoidal iput Frequecy trasfer fuctio Bode plot Examples Accelerometer Vehicle suspesio